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07 Oct 2009, 05:11

Hi everyone, I read the explanation for this question but I'm not convinced how if the angle ABC is 90, then we can conclude that \(a^2=\frac{1}{a}\)

According to me if the angle ABC is 90, then the triangle becomes isosceles but the sides are \(a^2\) and \(\frac{2}{a}\). Because is a triangle rectangle \((\frac{2}{a})^2 = (a^2)^2 + (a^2)^2\) ---> \(2 = a^6\)

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07 Oct 2009, 06:45

I'm sorry I tried to look for that thread but couldn't find it. The normal search does not work properly, and I cannot access to the other thread you post. I'll avoid duplicating threads next time. sorry for the inconvenience.

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07 Oct 2009, 07:03

Sorry, there was a typo. You must be able to access both threads now.

mikeCoolBoy wrote:

I'm sorry I tried to look for that thread but couldn't find it. The normal search does not work properly, and I cannot access to the other thread you post. I'll avoid duplicating threads next time. sorry for the inconvenience.